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Find the area of the region bounded by the curves $y=6-x^{2}$ and $y=x+4$.
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$\textbf{Answer:}$

$A= - \frac{9}{2}$

$\textbf{Explanation:}$

Let $y_1(x) = 6-x^2$ and $y_2(x) = x+4$

Find the end points of the area, for the end points, we have that;

$y_1(x) =y_2(x)$

$6-x^2 = x+4$

$x^2 +x-2=0$

$(x-1)(x+2)=0$

$\implies x=1$ and $x=-2$

$\implies a=-2$ and $b=1$

To see which of the functions is greater, we need to evaluate them at a point between $-2$ and $1$, lets choose $0$

$y_1(0) = 6-(0)^2 = 6$

$y_2(0) = 0+4=4$

$y_1(x)$ is the greater one, hence we have

$A = \int^b_a (y_1(x) -y_2(x)) dx$

$A = \int^1_{-2} ((6-x^2)-(x+4))dx$

$A = \int^1_{-2} (x^2+x-2 )dx$

$A = \frac{1}{3}x^3|^1_{-2}+\frac{1}{2}x^2|^1_{-2}-2x|^1_{-2}$

$A = 3- \frac{3}{2} +6$
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